Question

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad / s). If a particular disk is spun at 558.2 rad / s while it is being read, and then is allowed to come to rest over 0.435 seconds, what is the magnitude of the average angular acceleration of the disk?


____rad/s^2




If the disk is 0.12 m in diameter, what is the magnitude of the linear acceleration of a point 1/11 of the way out from the center of the disk?


______m/s^2

Answer 1

Answer:

Explanation:

Given

initial angular velocity $\omega _0=558.2 rad/s$

Time taken to stop $t=0.435 s$

using kinematic relation

$\omega =\omega _0+\alpha \cdot t$

here $\omega =$final angualr velocity is zero because it is coming to rest

$\alpha =$angular acceleration

thus $\alpha =-\frac{\omega _0}{t}$

$\alpha =-\frac{558.2}{0.435}=-1283.21 rad/s^2$

Magnitude is 1283.21 rad/s

If the diameter of disk $d=0.12 m$

then

radius $r=0.06 m$

Linear acceleration is givenby $\alpha \times r$

here radial distance is $\frac{1}{11}$ from center

thus $r'=\frac{r}{11}=\frac{0.06}{11}=0.00545$

$a_L=1283.21\times 0.00545=6.99 m/s^2$

             

Answer 2

Answer:

Explanation:

initial angular velocity, ωo = 558.2 rad/s

final angular velocity, ω = 0

time, t = 0.435 s

Use first equation of motion

ω = ωo + αt

where, α is the angular acceleration

0 = 558.2 + α x 0.435

α = - 1283.22 rad/s²

diameter of disc = 0.12 m

radius of disc, r = 0.06 m

distance, d = 0.06 / 11 m

linear acceleration, a = d x α = - 0.06 x 1283.22 / 11

a = - 7 m/s²